Fusion — solar fusion, tokamak (concept): Visual walkthrough

We assume you know nothing except that atoms have a tiny central lump (the nucleus) made of protons and neutrons. Everything else — mass defect, binding energy, $c^{2}$ , the units — we earn as we go.

Step 1 — What a nucleus is made of, and what "mass" means here

WHAT. A nucleus is a bag of two kinds of particle:

a proton (positive charge, symbol $p$ ),
a neutron (no charge, symbol $n$ ).

We call any nucleon (proton or neutron) a nucleon. The reaction has four players:

Symbol	Name	Nucleons inside
$_{1}^{2} D$	deuterium (heavy hydrogen)	1 proton + 1 neutron
$_{1}^{3} T$	tritium (heavier hydrogen)	1 proton + 2 neutrons
$_{2}^{4} He$	helium-4	2 protons + 2 neutrons
$n$	a lone neutron	1 neutron

The little numbers: the top number is the total nucleon count $A$ ; the bottom number is the proton count $Z$ .

WHY we care about mass. Every particle has a rest mass — a "weight" it has even when sitting still. We measure it in atomic mass units (u), where 1 u is roughly the mass of one proton. Mass matters because — as we'll see — mass is a storage tank for energy.

PICTURE. Count the balls on each side. Notice both sides have the same total: 2 protons and 3 neutrons. Nucleons are just rearranged, not created or destroyed.

Step 2 — The masses don't add up (that's the whole point)

WHAT. Let's literally weigh both sides using the given rest masses (in u):

$2.014102 m_{D} + 3.016049 m_{T} vs. 4.002603 m_{H e} + 1.008665 m_{n}$

Before: $2.014102 + 3.016049 = 5.030151 u$
After: $4.002603 + 1.008665 = 5.011268 u$

The after pile is lighter, even though it has the exact same 5 nucleons!

WHY this happens. The helium nucleus holds its nucleons together far more tightly than the loose D and T did. Tightly-bound stuff is lighter — binding "uses up" mass. (We make this precise in Step 4.)

PICTURE. Two pans of a balance. Left pan (reactants) tips down — it's heavier. The tiny sliver that's missing on the right is labelled $Δ m$ .

Why subtract inputs minus outputs and not the other way? Because we want the loss to come out positive, so that a positive $Δ m$ means "mass was released."

Step 3 — Why does missing mass mean energy? ( $E = m c^{2}$ )

WHAT. Einstein's rule says rest mass and energy are the same thing in different clothes: $E = m c^{2}$

$E$ — energy (in joules if $m$ is in kg),
$m$ — mass,
$c$ — the speed of light, $c = 3 \times 1 0^{8} m/s$ ,
$c^{2}$ — that speed squared, a colossal $9 \times 1 0^{16}$ .

WHY $c^{2}$ and not just $c$ ? $c^{2}$ is the exchange rate between the currency of mass and the currency of energy. Because $c^{2}$ is enormous, even a speck of mass converts into a mountain of energy. This is the tool that answers our question "the mass vanished — where did it go?" — it went into energy, at rate $c^{2}$ .

PICTURE. A pinhead of mass on the left; an arrow through the "× $c^{2}$ " machine; an explosion of energy on the right. Same object, blown up by $9 \times 1 0^{16}$ .

See Mass-Energy Equivalence (E=mc^2) for the full story of this relation.

Step 4 — The binding-energy curve: why helium is lighter

WHAT. For every nucleus, define its binding energy per nucleon, written $B / A$ :

$B$ = energy you'd have to pump in to rip the nucleus completely apart into free nucleons,
$A$ = number of nucleons,
$B / A$ = binding energy shared out per nucleon — a fairness-adjusted measure of "how tightly is each nucleon held?"

Plot $B / A$ against $A$ and you get the famous curve: it rises steeply from hydrogen, peaks at iron ( $^{56} Fe$ , about 8.8 MeV per nucleon), then gently falls.

WHY this decides everything. More binding energy $B$ means more mass was "used up" to build the nucleus (by $E = m c^{2}$ ), so the nucleus is lighter per nucleon. Fusing D and T into He moves us up the steep left slope — helium sits higher, is more tightly bound, and is therefore lighter. The height you climbed on this curve is exactly the energy released.

PICTURE. The curve, with D and T marked low on the steep slope, and He-4 marked much higher. A vertical green arrow shows the "climb" — that climb is the energy $Q$ . Note fusion only pays off left of iron; right of iron you must go the other way (Nuclear Fission).

Step 5 — The unit trick: turning "u" into MeV

WHAT. We have $Δ m = 0.018883 u$ . We want an energy in useful nuclear units. Physicists measure nuclear energy in MeV (mega-electron-volts). There's a ready-made conversion: $1 u \cdot c^{2} = 931.5 MeV$

WHY this shortcut exists. Rather than convert u → kg, multiply by $c^{2}$ in joules, then convert joules → MeV every single time, physicists baked the whole chain into one number: 931.5 MeV per u. It already contains the $c^{2}$ . So: $Q = Δ m \times 931.5 \frac{MeV}{u}$

PICTURE. A conversion "ruler": on the left, mass in u; on the right, energy in MeV; the peg linking them is 931.5.

Step 6 — Put it together: the 17.6 MeV appears

WHAT. Multiply: $Q = 0.018883 u \times 931.5 \frac{MeV}{u} = 17.59 MeV \approx 17.6 MeV$

WHY this number is special. 17.6 MeV per single D–T fusion is huge per particle — millions of times more than any chemical reaction (which yields a few eV per atom). This is why fusion is so tantalising as a power source.

PICTURE. A bar-chart of energy: the tall D–T bar (17.6 MeV) towers over a chemical-reaction bar (a few eV, essentially invisible next to it).

Step 7 — The edge case: what if $Δ m$ came out negative?

WHAT. Suppose we tried to "fuse" two nuclei heavier than iron. Then the product sits lower on the $B / A$ curve, is less tightly bound, and is therefore heavier than the reactants. Now $Δ m = m_{in} - m_{out} < 0 \Rightarrow Q < 0.$

WHY it matters. A negative $Q$ means the reaction absorbs energy instead of releasing it — you'd have to push energy in, and it would never run on its own. This is the boundary that separates fusion-that-works (light nuclei) from fusion-that-fails (heavy nuclei).

PICTURE. The same curve as Step 4, but now starting on the right (past iron) and trying to go up by fusing — you'd fall down the slope instead. Red arrow pointing the wrong way, labelled " $Q < 0$ : costs energy."

Step 8 — The other edge case: energy vanishes if binding doesn't change

WHAT. Imagine a hypothetical "reaction" where the product is bound exactly as tightly as the reactants — same total $B / A$ . Then nothing is used up: $m_{in} = m_{out}$ , so $Δ m = 0 \Rightarrow Q = 0.$

WHY it matters. This is the break-even line of the physics: at the peak of the curve (iron), the slope flattens, $B / A$ barely changes, and $Q \to 0$ . Iron is the "ash" of the nuclear world — you can't squeeze energy out of it in either direction. Neither fusion nor fission pays near iron.

PICTURE. Zoom on the flat top of the curve near iron: two points at the same height, connected by a flat arrow labelled " $Δ m = 0$ , $Q = 0$ ."

The one-picture summary

Everything above, compressed into a single flow: mass in → mass out → the missing sliver $Δ m$ → ×931.5 → 17.6 MeV, with the binding-energy climb drawn beside it as the reason.

Recall Feynman retelling of the whole walkthrough

Take one heavy-hydrogen bit (deuterium) and one super-heavy-hydrogen bit (tritium). Weigh them both — write the number down. Now let them stick together into a helium nucleus plus a spare neutron. Weigh that. Surprise: it weighs a tiny bit less, even though not a single ball was lost! Where did the weight go? It became energy, at the crazy exchange rate $c^{2}$ . The reason the helium is lighter is that its nucleons hug each other much harder — and hugging harder "spends" mass. On the binding-energy curve, we climbed a hill from D and T up to helium, and the height of that climb is the energy that came out: about 17.6 million electron-volts. If instead we'd tried to fuse two nuclei heavier than iron, we'd be sliding down the hill — that costs energy, so it never happens on its own. And right at the top of the hill (iron), the ground is flat, so nothing comes out at all. That's the entire story of why the Sun shines and why we chase tokamaks: light bits love to lump, and lumping pays.

Recall Quick self-check

What sign must $Δ m$ have for fusion to release energy? ::: Positive — the products must be lighter than the reactants. What single number converts a mass loss in u directly to MeV? ::: 931.5 MeV per u. Why is helium-4 lighter than the D and T that made it? ::: It is more tightly bound; that stronger binding "used up" some rest mass. What does the height climbed on the $B / A$ curve represent physically? ::: The energy $Q$ released in the reaction.